$g(x)=3x^2+x$ What is the average rate of change of $g$ over the interval $-3\leq x \leq -3+h$, in terms of $h$, where $h\neq 0$ ? Your answer must be fully expanded and simplified.
Answer: This is the formula for the average rate of change of a function $f$ over the interval $[a,b]$ : $\dfrac{f(b)-f(a)}{b-a}$ We can calculate that $g(-3)=24$. We are interested in the average rate of change of $g(x)=3x^2+x$ over the interval $-3\leq x \leq -3+h$ : $\begin{aligned} &\phantom{=}\dfrac{g(-3+h)-g(-3)}{(-3+h)-(-3)} \\\\ &=\dfrac{3(-3+h)^2+(-3+h)-24}{-3+h-(-3)} \\\\ &=\dfrac{3(-3+h)^2+(-3+h)-24}{h} \end{aligned}$ We can now simplify the expression we obtained. $\begin{aligned} &\phantom{=}\dfrac{3(-3+h)^2+(-3+h)-24}{h} \\\\ &=\dfrac{3(9-6h+h^2)-3+h-24}{h} \\\\ &=\dfrac{27-18h+3h^2-3+h-24}{h} \\\\ &=\dfrac{3h^2-17h}{h} \\\\ &=\dfrac{h(3h-17)}{h} \\\\ &=3h-17\text{, for }h\neq 0 \end{aligned}$ Since we are given that $h\neq 0$, the average rate of change of the function is $3h-17$. Notice that the average rate of change is calculated just like the slope of the secant line that intersects the graph of the function at the interval's endpoints.